At first sight you might say that, since NaOH is 100% dissociated, then sf([OH^-]=10^(-7)color(white)(x)"mol/l").
Since sf(pH+pOH=14) this would mean that such a solution would be sf(pH=7) i. e neutral.
We would predict that such a very dilute solution of alkali would have a pH very slightly above 7.
At such a low concentration as this, we must also take into account the ions present from the dissociation of water:
sf(H_2OrightleftharpoonsH^(+)+OH^(-))
For which sf(K_w=[H^+][OH^-]=10^(-14)" ""at"" "25^@C)
and sf([H^+]=10^(-7)M) and sf([OH^-]=10^(-7)M).
To get the total sf(OH^-) concentration you might think that this would therefore be sf(10^(-7)+10^(-7)=2xx10^(-7)color(white)(x)"mol/l").
This is not the case. This represents the initial conditions of an equilibrium system which has been disturbed.
The reaction quotient is now sf(10^(-7)xx2xx10^(-7)=2xx10^(-14)) which exceeds sf(K_w).
Le Chatelier's Principle tells us that the system will relax to restore the postion of equilibrium and the value of sf(K_w).
It will do this by shifting to the left, for which we must set up an ICE table based on sf("mol/l"):
sf(color(white)(xxxx)H_2Ocolor(white)(xxxxx)rightleftharpoonscolor(white)(xxxxx)H^(+)color(white)(xxxxxx)+color(white)(xxxxx)OH^-)
sf(Icolor(white)(xxxxxxxxxxxxxxxxxx)10^(-7)color(white)(xxxxxxxxx)(10^(-7)+10^(-7)))
sf(Ccolor(white)(xxxxxxxxxxxxxxxxx)-xcolor(white)(xxxxxxxxxxxxxx)-x)
sf(Ecolor(white)(xxxxxxxxxxxxxxx)(10^(-7)-x)color(white)(xxxxxxxx)(2xx10^(-7)-x))
:.sf((10^(-7)-x)(2xx10^(-7)-x)=10^(-14))
This becomes:
sf(x^(2)-(3xx10^(-7))x+10^(-14)=0)
If we apply the quadratic formula we get the 2 roots:
sf(x=2.615xx10^(-7)color(white)(x)"mol/l")
or
sf(x=0.385xx10^(-7)color(white)(x)"mol/l")
I will discard the 1st because this will give a -ve concentration so:
sf([OH^-]=2xx10^(-7)-0.385xx10^(-7)=1.615xx10^(-7)color(white)(x)"mol/l")
To get the pH you can say:
sf(pOH=-log[OH^(-)]=-log[1.615xx10^(-7)]=6.791)
sf(pH=14-pOH=14-6.791=7.21)
Which is the type of result predicted.
